Lecture 8 1. Associated Primes and Primary decomposition for Modules We would like to describe in more detail the structure of submodules of a finitely generated module M over a Noetherian ring A. We will do this by investigating special submodules (called primary) and special primes of A called associated primes. Throughout this lecture, A will be a ring and M, N etc will be A-modules. Definition 1.1. Let a A. Then θ a : M M defined by θ a (m) = am, for m M is called the homothety of ration a. Definition 1.2. Let A Noetherian. We say that M is coprimary if for all a A, θ a is either injective or nilpotent. Proposition 1.3. If M is coprimary, then is a prime ideal. {a : θ a is nilpotent} = Ann A (M) Proof. Let ab Ann A (M) and assume that a Ann A (M). So, θ a is not nilpotent, hence it is injective. That is, am = 0 implies m = 0. But, since there exists n such that (ab) n M = 0, we get b n a n M = 0 and, hence, a n M = 0 or a Ann A (M). Definition 1.4. If M is coprimary and if P = {a : θ a is nilpotent}, then we say that M is P -coprimary. Proposition 1.5. Let M be a P -coprimary finitely generated A-module and 0 N a submodule of M. Then N is P -coprimary. Proof. It follows directly from the definition that N is coprimary. AnnA (N = P. Let us prove that Since N M then Ann A (M) Ann A (N) and then P = Ann A (M) Ann A (N). 1
2 Now let a Ann A (N) and n minimal such that a n N = 0. Since a n 1 N 0 this means that θ a is not injective on M. Hence it is nilpotent on M and so a P. Theorem 1.6. Let M be a P -coprimary module. Let Q be a prime ideal of A. Then A/Q injects into M as an A-submodule, i.e. there exists an A-linear injection A/Q M, if and only if Q = P. Proof. First let us who that we can embed A/P into M (via an A-linear map). It is enough to show that A/P embeds into a cyclic submodule of M. Now, let us note that every cyclic submodule Am with m 0 is P -coprimary by the above result. So we can assume that M = Am, for some m 0. Let I = Ann A (m). Then A M via φ(a) = am has kernel equal to I and so A/I is isomorphic to Am = M. We will show that A/P can be embedded in A/I. Let n minimal with the property P n I. This n exists because A/I is P -coprimary. Let x P n 1 \ I. Consider the map f : A A/I f(x) = ax, for a A. Let us prove that Ker(f) = P. Since P x P n I, we see that P is in the kernel. Now let a such that ax I. Then axm = 0. But xm 0 as x I. Then θ a is not injective on M and so θ a must be nilpotent. In other words, a P. Therefore A/P A/I via f, and this shows that A/P embbeds into M. Now assume that A/Q embeds into M, then A/Q is P -coprimary. But Ann ( A/Q) = Q is prime and hence Q = Q = P. Definition 1.7. Let N M. Then N is called primary if for all a A, m M such that am N and m / N, then n such that a n M N. In other words, if a is a zerodivisor of M/N, then a Rad(Ann A (M/N)). Remark 1.8. I A is a primary ideal if and only if I is a primary submodule of A.
Example 1.9. In A = k[x, y], consider (xy, x 2 ) = (x) (x 2, y) where the first ideal is prime and the latter is primary. We can also write (xy, x 2 ) = (x) (x, y) 2 = (x) (x 2, xy, y 2 ). Note that (x 2, y) = (x 2, xy, y 2 ) = (x, y). As one can note, the number of terms in the decompositions as well as the radicals of the ideals that are part of the decompositions are the same. This is no coincidence, and the primary decomposition theory will explains this phenomenon. Definition 1.10. If A is a ring, M is an A-module, and P Spec(A). Then P is called an associated prime if 0 m M such that P = Ann A (m). The set of associated primes is denoted Ass A (M). Remark 1.11. Am A = Ann A (m). If Ann A(m) = P, then Am = A/P M since Am = A m M. On the other hand, if A/P M, then Ann A (1) = P, and A 1 M, hence P Ass(M). So P Ass(M) A/P M. Proposition 1.12. Take P Ass(M). Denote F = Am = A/P M, and P = Ann A (m). If 0 y F, then Ann A (y) = P. 3 Proof. We know y = am, so P Ann A (y). Now take b A such that by = 0. This implies abm = 0, so ab P, and since a / P, we must have b P. Thus Ann A (y) P. Remark 1.13. If M = M i, then Ass(M) = Ass(M i ). i I i I Definition 1.14. Let M be a module over A. If Ass(M) = 1, then M is called A- coprimary. Proposition 1.15. Take M an A-module such that M 0. Let P = {Ann A (m) 0 m M}. The maximal elements of P are prime. Proof. Let P be such a maximal element. Let ab P, so abm = 0. We know bm 0 if and only if b / P, so Ann(m) Ann(bm). But P = Ann(m) is maximal, so P = Ann(m) = Ann(bm). Since a Ann(bm), we have a P, and have shown that P is prime. Corollary 1.16. If 0 A M with A Noetherian, then Ass(M). Moreover, every zerodivisor of M lives in some associated prime of M. More precisely, ZD(M) = {a A m 0, m M am = 0} = P.
4 Proof. For the first part, if A is Noetherian, then P has maximal elements. For the second, let a be a zerodivisor of M. So am = 0 for some m 0, so a Ann A (m). But for all ideals I = Ann A (m), one can find a maximal element of P that contains I. By the previous theorem, this element is an associated prime, say Q. So a Ann(m) Q Ass(M). So ZD(M) P. For the reverse implication, if a P then a P Ass(M), so P = Ann A (m) for some nonzero m M. But then am = 0 which implies a is a zerodivisor of M. Proposition 1.17. If 0 M M M 0 is a short exact sequence, then Ass(M ) Ass(M ) Ass(M ). Proof. Denote by µ : M M and σ : M M the maps that appear in our short exact sequence. For the first inclusion, if P Ass(M ) then 0 m such that P = Ann(m ). But P = Ann(µ(m )) since µ is injective, and µ(m ) 0. So P Ass(M). For the second inclusion, let P Ass(M). Then P = Ann(m) for some 0 m M. If Am Im(µ) 0, then m 0 = µ(m 0) = am 0, with m 0 M. Remember that P = Ann(m 0 ) = Ann(µ(m 0)) = Ann(m 0) since µ is injective. Thus P Ass(M ). If Am Im(µ) = 0, then P = Ann(σ(m)). If a P then am = 0 which implies aσ(m) = 0 σ(am) = 0, so am ker σ = Im(µ), so am Im(µ), so am = 0 which gives a P. Also, σ(m) 0, since if not, then m ker σ = Im(µ) which says m Am Im(µ) = 0, a contradiction. n Corollary 1.18. Ass( M i ) = Ass(M i ). i=1 i=1 Proof. We show for n = 2, which is the main case that implies the general statement. Consider the exact sequence 0 M 1 M 1 M 2 M 2 0. By the above theorem, we have that both Ass(M 1 ) and Ass(M 2 ) belong to Ass(M 1 M 2 ) Ass(M 1 ) Ass(M 2 ). 2. The Prime Filtration Proposition 2.1. Let M be finitely generated over A, and A be a Noetherian ring. Then there is a chain of submodules of M 0 = M 0 M 1 M n = M with M i /M i 1 = A/Pi where P i Spec(A).
Proof. We proceed by induction on n. The n = 0 case is trivial. Assume we have chosen M 0,..., M i 1. If M i 1 M, then M/M i 1 0, so P i Ass(M/M i 1 ) Spec(A). Recall that A/P i M/M i 1. Take M i M such that M i /M i 1 = A/P i which is the image of the inclusion. So we have M i as needed, and the procedure stops at M n = M since M is Noetherian. 5 The above result often allows one to reduce statements about finitely generated arbritary A-modules to modules of the type A/P, P prime ideal, by using techinques involving short exact sequences whenever possible. Corollary 2.2. If A is Noetherian, and M is finitely generated over A, then Ass(M) is finite. Proof. Look at our prime filtration 0 = M 0 M 1 M n = M. Then Ass(M) Ass(M n 1 ) Ass(M/M n 1 ) since as we have the following short exact sequence 0 M n 1 M M/M n 1 0. Repeat for M n 1, and obtain that Ass(M) Ass(M i /M i 1 ) = Ass(A/P i ) = {P 1,..., P n }. i=1 Remark 2.3. Under the conditions of the Proposition we have that Ass(M) {P 1,.., P n } as the proof of the Corollary shows. i=1 If A M is Noetherian and Artinian, then by definition one can produce a filtration (called composition series) as in the Proposition 2.1 where the factors are simple modules, and hence quotients of A by maximal ideals. The important fact is that Proposition 2.1 holds for a much larger class of modules.